what factors would cause the percent yield to be low
The earth of pharmaceutical product is an expensive ane. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop meliorate ways to make drugs faster and more efficiently. Studying how much of a chemical compound is produced in whatsoever given reaction is an important office of cost command.
Percent Yield
Chemical reactions in the real globe don't always get exactly equally planned on paper. In the grade of an experiment, many things will contribute to the formation of less product than predicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the per centum yield.
To compute the percent yield, it is first necessary to determine how much of the production should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that can exist formed from the given amounts of reactants. The actual yield is the amount of production that is actually formed when the reaction is carried out in the laboratory. The pct yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
\[\text{Percentage Yield} = \frac{\text{Bodily Yield}}{\text{Theoretical Yield}} \times 100\%\]
Percent yield is very important in the industry of products. Much time and money is spent improving the per centum yield for chemic production. When circuitous chemicals are synthesized by many different reactions, one pace with a low pct yield tin can apace cause a big waste of reactants and unnecessary expense.
Typically, percent yields are understandably less than \(100\%\) because of the reasons indicated earlier. Even so, per centum yields greater than \(100\%\) are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemic, he or she is always careful to purify the products of the reaction. Example \(\PageIndex{1}\) illustrates the steps for determining percent yield.
Example \(\PageIndex{1}\): Decomposition of Potassium Chlorate
Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction beneath:
\[ii \ce{KClO_3} \left( southward \correct) \rightarrow 2 \ce{KCl} \left( s \correct) + 3 \ce{O_2} \left( g \right)\nonumber\]
In a sure experiment, \(40.0 \: \text{g} \: \ce{KClO_3}\) is heated until it completely decomposes. The experiment is performed and the oxygen gas is collected and its mass is institute to be \(fourteen.9 \: \text{1000}\).
- What is the theoretical yield of oxygen gas?
- What is the percent yield for the reaction?
Solution
a. Calculation of theoretical yield
First, nosotros will summate the theoretical yield based on the stoichiometry.
Step one: Identify the "given" data and what the problem is request y'all to "find".
Given: Mass of \(\ce{KClO_3} = 40.0 \: \text{g}\)
Mass of Otwo nerveless = fourteen.9g
Discover: Theoretical yield, g O2
Pace 2: List other known quantities and programme the trouble.
i mol KClOiii = 122.55 k/mol
one mol Oii - 32.00 chiliad/mol
Step three: Apply stoichiometry to convert from the mass of a reactant to the mass of a product:
Step 4: Solve.
\[xl.0 \: \cancel{\text{g} \: \ce{KClO_3}} \times \frac{one \: \cancel{\text{mol} \: \ce{KClO_3}}}{122.55 \: \cancel{\text{thou} \: \ce{KClO_3}}} \times \frac{3 \: \cancel{\text{mol} \: \ce{O_2}}}{2 \: \cancel{\text{mol} \: \ce{KClO_3}}} \times \frac{32.00 \: \text{g} \: \ce{O_2}}{one \: \abolish{\text{mol} \: \ce{O_2}}} = xv.7 \: \text{g} \: \ce{O_2}\nonumber\]
The theoretical yield of \(\ce{O_2}\) is \(15.7 \: \text{k}\), 15.half-dozen7 chiliad unrounded.
Step 5: Think near your result.
The mass of oxygen gas must exist less than the \(40.0 \: \text{g}\) of potassium chlorate that was decomposed.
b. Calculation of per centum yield
Now nosotros will use the actual yield and the theoretical yield to calculate the percentage yield.
Pace 1: Identify the "given" data and what the problem is asking yous to "find".
Given: Theoretical yield =fifteen.6seven g, use the united nations-rounded number for the calculation.
Bodily yield = 14.9g
Notice: Percent yield, % Yield
Stride two: List other known quantities and plan the problem.
No other quantities needed.
Footstep 3: Use the percent yield equation below.
\(\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\)
Step 4: Solve.
\(\text{Percent Yield} = \frac{14.nine \: \text{g}}{15.\underline{six}7 \: \text{g}} \times 100\% = 94.9\%\)
Step 5: Think about your result.
Since the actual yield is slightly less than the theoretical yield, the percent yield is just under \(100\%\).
Example \(\PageIndex{2}\): Oxidation of Zinc
Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained co-ordinate to the equation:
\(\ce{CuSO4}(aq)+\ce{Zn}(s)\rightarrow \ce{Cu}(south)+\ce{ZnSO4}(aq)\)
What is the per centum yield?
Solution
| Steps for Problem Solving-The Product Method | Case \(\PageIndex{1}\) |
|---|---|
| Identify the "given" information and what the trouble is asking you to "notice." | Given: 1.274 g CuSO4 Actual yield = 0.392 g Cu |
| Listing other known quantities. | ane mol CuSO4= 159.62 yard/mol Since the amount of product in grams is non required, simply the tooth mass of the reactants is needed. |
| Residuum the equation. | The chemical equation is already balanced. The counterbalanced equation provides the relationship of 1 mol CuSO4 to 1 mol Zn to 1 mol Cu to ane mol ZnSO4. |
| Prepare a concept map and employ the proper conversion factor. | The provided information identifies copper sulfate as the limiting reactant, then the theoretical yield (g Cu) is found past performing mass-mass calculation based on the initial amount of CuSO4. |
| Cancel units and calculate. | \[\mathrm{ane.274\:\cancel{grand\:Cu_SO_4}\times \dfrac{1\:\cancel{mol\:CuSO_4}}{159.62\:\cancel{g\:CuSO_4}}\times \dfrac{1\:\cancel{mol\: Cu}}{1\:\abolish{mol\:CuSO_4}}\times \dfrac{63.55\:thousand\: Cu}{1\:\cancel{mol\: Cu}}=0.5072\: g\: Cu}\nonumber\] Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be: \[\mathrm{percent\: yield=\left(\dfrac{actual\: yield}{theoretical\: yield}\right)\times 100}\] \[\begin{marshal} |
| Call up about your result. | Since the actual yield is slightly less than the theoretical yield, the percent yield is simply nether \(100\%\). |
Exercise \(\PageIndex{1}\)
What is the pct yield of a reaction that produces 12.5 g of the Freon CF2Cl2 from 32.ix 1000 of CClfour and excess HF?
\[\ce{CCl4 + 2HF \rightarrow CF2Cl2 + 2HCl} \nonumber\]
- Answer
- 48.3%
Summary
Theoretical yield is calculated based on the stoichiometry of the chemic equation. The bodily yield is experimentally determined. The percent yield is determined past computing the ratio of actual yield to theoretical yield.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development squad to meet platform mode, presentation, and quality:
-
Marisa Alviar-Agnew (Sacramento City Higher)
-
Henry Agnew (UC Davis)
Source: https://chem.libretexts.org/Courses/Sacramento_City_College/SCC:_CHEM_300_-_Beginning_Chemistry/SCC:_CHEM_300_-_Beginning_Chemistry_%28Faculty%29/08:_Quantities_in_Chemical_Reactions/8.06:_Limiting_Reactant,_Theoretical_Yield,_and_Percent_Yield_from_Initial_Masses_of_Reactants
0 Response to "what factors would cause the percent yield to be low"
Enregistrer un commentaire